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## Credits

Vishwakarma Government Engineering College (VGEC), Ahmedabad
Urvi Rathod has created this Calculator and 1000+ more calculators!
Osmania University (OU), Hyderabad
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## Load Current Using Area Of X-Section(1-Phase 2-Wire OS) Solution

STEP 0: Pre-Calculation Summary
Formula Used
current4 = sqrt(Area Of 1-Φ 2-wire system*Line Losses/(Resistivity*Length))
I4 = sqrt(a4*W/(ρ*l))
This formula uses 1 Functions, 4 Variables
Functions Used
sqrt - Squre root function, sqrt(Number)
Variables Used
Area Of 1-Φ 2-wire system - The Area Of 1-Φ 2-wire system is the amount of two-dimensional space taken up by an object. (Measured in Square Meter)
Line Losses - Line Losses is defined as the losses that are produced in the line. (Measured in Watt)
Resistivity - Resistivity is the measure of how strongly a material opposes the flow of current through them. (Measured in Ohm Meter)
Length - Length is the measurement or extent of something from end to end. (Measured in Meter)
STEP 1: Convert Input(s) to Base Unit
Area Of 1-Φ 2-wire system: 6 Square Meter --> 6 Square Meter No Conversion Required
Line Losses: 0.6 Watt --> 0.6 Watt No Conversion Required
Resistivity: 1.7E-05 Ohm Meter --> 1.7E-05 Ohm Meter No Conversion Required
Length: 3 Meter --> 3 Meter No Conversion Required
STEP 2: Evaluate Formula
Substituting Input Values in Formula
I4 = sqrt(a4*W/(ρ*l)) --> sqrt(6*0.6/(1.7E-05*3))
Evaluating ... ...
I4 = 265.684465662029
STEP 3: Convert Result to Output's Unit
265.684465662029 Ampere --> No Conversion Required
FINAL ANSWER
265.684465662029 Ampere <-- Current Of 1-Φ 2-wire system
(Calculation completed in 00.031 seconds)

## < 8 Area Of X-Section Calculators

Maximum Voltage Using Area Of X-section(1-phase 2-wire OS)
maximum_voltage = sqrt((4*Length*Resistivity*(Power Transmitted^2))/(Area Of 1-Φ 2-wire system*Line Losses*((cos(Theta))^2))) Go
RMS Voltage Using Area Of X-Section(1-Phase 2-Wire OS)
rms_voltage = sqrt((2*Length*Resistivity*(Power Transmitted^2))/(Area Of 1-Φ 2-wire system*Line Losses*((cos(Theta))^2))) Go
Power Transmitted Using Area Of X-section(1-phase 2-wire OS)
power_transmitted = sqrt((Area Of X-Section*(Maximum Voltage^2)*Line Losses*((cos(Theta))^2))/(4*Resistivity*Length)) Go
Power Factor Using Area Of X-section(1-phase 2-wire OS)
power_factor = sqrt((4*Power Transmitted^2)*Resistivity*Length/(Area Of 1-Φ 2-wire system*Line Losses*(Maximum Voltage^2))) Go
Line Losses Using Area Of X-section(1-phase 2-wire OS)
line_losses = (4*Length*Resistivity*(Power Transmitted^2))/(Area Of 1-Φ 2-wire system*(Maximum Voltage^2)*((cos(Theta))^2)) Go
Length Of Wire Using Area Of X-section(1-phase 2-wire OS)
length = Area Of 1-Φ 2-wire system*(Maximum Voltage^2)*Line Losses*((cos(Theta))^2)/(4*Resistivity*(Power Transmitted^2)) Go
Resistivity Using Area Of X-section(1-phase 2-wire OS)
resistivity = Area Of 1-Φ 2-wire system*(Maximum Voltage^2)*Line Losses*((cos(Theta))^2)/(4*Length*(Power Transmitted^2)) Go
Load Current Using Area Of X-Section(1-Phase 2-Wire OS)
current4 = sqrt(Area Of 1-Φ 2-wire system*Line Losses/(Resistivity*Length)) Go

### Load Current Using Area Of X-Section(1-Phase 2-Wire OS) Formula

current4 = sqrt(Area Of 1-Φ 2-wire system*Line Losses/(Resistivity*Length))
I4 = sqrt(a4*W/(ρ*l))

## What is the value of maximum voltage and volume of conductor material in 1-phase 2-wire system?

The volume of conductor material required in this system is 2/cos2θ times that of 2-wire d.c.system with the one conductor earthed. The maximum voltage between conductors is vm so that r.m.s. value of voltage between them is vm/√2.

## How to Calculate Load Current Using Area Of X-Section(1-Phase 2-Wire OS)?

Load Current Using Area Of X-Section(1-Phase 2-Wire OS) calculator uses current4 = sqrt(Area Of 1-Φ 2-wire system*Line Losses/(Resistivity*Length)) to calculate the Current Of 1-Φ 2-wire system, The Load Current Using Area Of X-section(1-Phase 2-Wire OS) formula is defined as the current that flows into the load of the one-phase two-wire overhead system. Current Of 1-Φ 2-wire system and is denoted by I4 symbol.

How to calculate Load Current Using Area Of X-Section(1-Phase 2-Wire OS) using this online calculator? To use this online calculator for Load Current Using Area Of X-Section(1-Phase 2-Wire OS), enter Area Of 1-Φ 2-wire system (a4), Line Losses (W), Resistivity (ρ) and Length (l) and hit the calculate button. Here is how the Load Current Using Area Of X-Section(1-Phase 2-Wire OS) calculation can be explained with given input values -> 265.6845 = sqrt(6*0.6/(1.7E-05*3)).

### FAQ

What is Load Current Using Area Of X-Section(1-Phase 2-Wire OS)?
The Load Current Using Area Of X-section(1-Phase 2-Wire OS) formula is defined as the current that flows into the load of the one-phase two-wire overhead system and is represented as I4 = sqrt(a4*W/(ρ*l)) or current4 = sqrt(Area Of 1-Φ 2-wire system*Line Losses/(Resistivity*Length)). The Area Of 1-Φ 2-wire system is the amount of two-dimensional space taken up by an object, Line Losses is defined as the losses that are produced in the line, Resistivity is the measure of how strongly a material opposes the flow of current through them and Length is the measurement or extent of something from end to end.
How to calculate Load Current Using Area Of X-Section(1-Phase 2-Wire OS)?
The Load Current Using Area Of X-section(1-Phase 2-Wire OS) formula is defined as the current that flows into the load of the one-phase two-wire overhead system is calculated using current4 = sqrt(Area Of 1-Φ 2-wire system*Line Losses/(Resistivity*Length)). To calculate Load Current Using Area Of X-Section(1-Phase 2-Wire OS), you need Area Of 1-Φ 2-wire system (a4), Line Losses (W), Resistivity (ρ) and Length (l). With our tool, you need to enter the respective value for Area Of 1-Φ 2-wire system, Line Losses, Resistivity and Length and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.
How many ways are there to calculate Current Of 1-Φ 2-wire system?
In this formula, Current Of 1-Φ 2-wire system uses Area Of 1-Φ 2-wire system, Line Losses, Resistivity and Length. We can use 8 other way(s) to calculate the same, which is/are as follows -
• length = Area Of 1-Φ 2-wire system*(Maximum Voltage^2)*Line Losses*((cos(Theta))^2)/(4*Resistivity*(Power Transmitted^2))
• resistivity = Area Of 1-Φ 2-wire system*(Maximum Voltage^2)*Line Losses*((cos(Theta))^2)/(4*Length*(Power Transmitted^2))
• power_transmitted = sqrt((Area Of X-Section*(Maximum Voltage^2)*Line Losses*((cos(Theta))^2))/(4*Resistivity*Length))
• line_losses = (4*Length*Resistivity*(Power Transmitted^2))/(Area Of 1-Φ 2-wire system*(Maximum Voltage^2)*((cos(Theta))^2))
• maximum_voltage = sqrt((4*Length*Resistivity*(Power Transmitted^2))/(Area Of 1-Φ 2-wire system*Line Losses*((cos(Theta))^2)))
• power_factor = sqrt((4*Power Transmitted^2)*Resistivity*Length/(Area Of 1-Φ 2-wire system*Line Losses*(Maximum Voltage^2)))
• rms_voltage = sqrt((2*Length*Resistivity*(Power Transmitted^2))/(Area Of 1-Φ 2-wire system*Line Losses*((cos(Theta))^2)))
• current4 = sqrt(Area Of 1-Φ 2-wire system*Line Losses/(Resistivity*Length))
Where is the Load Current Using Area Of X-Section(1-Phase 2-Wire OS) calculator used?
Among many, Load Current Using Area Of X-Section(1-Phase 2-Wire OS) calculator is widely used in real life applications like {FormulaUses}. Here are few more real life examples -
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